JEE Advanced : MOCK TEST PAPER 2016

[Type text]

JPT-1 (JEE ADVANCE) TEST DATE : | BATCH : JP,JF,JR

S.No. Subject Nature of Questions No. of Questions Marks Negative Total

1 to 12 SCQ 12 3 �1 36

13 to 20 Comprehenstion (4 x 2Ques) 8 3 �1 24

21 to 25 Integer Type Questions (Two Digits Answer) 5 3 0 15

26 to 37 SCQ 12 3 0 36



51 to 62 SCQ 12 3 0 36



75 225

Paper-1

Total Total

Maths

Physics

Chemistry

1. Let A be a m × n matrix of distinct real numbers. If x is the minimum of maximums of each row and y is the

maximum of minimums of each column, then [MT-AL] [301]

(A) x > y (B) x = y

(C) x < y (D*) nothing can be said with surely

ekuk A, m × n Øe dk fHkUu-fHkUu okLrfod la[;kvksa ls cuk ,d vkO;wg gSA ;fn izR;sd iafDr ds vf/kdre ekuksa esa

U;wure eku x gS rFkk izR;sd LraEHk ds U;wure ekuksa esa vf/kdre eku y gS] rc

(A) x > y (B) x = y

(C) x < y (D*) okLro esa dqN ugha dgk tk ldrk gSA

2. In a game a coin is tossed (2n + m) times and a player wins if he does not get any two consecutive

outcomes same for atleast 2n times in a row. The probability that he win the game is - [PR-CD] [301]

fdlh [ksy esa ,d flDds dks (2n + m) ckj mNkyk tkrk gS rFkk ;fn ,d iafDr esa de ls de 2n ckj esa ;fn dksbZ

f[kykMh dksbZ Hkh nks laxr ifj.kke leku izkIr ugha djrk gS] rks og thr tkrk gSA izkf;drk Kkr d hft, fd og [ksy

dks thrrk gS -

(A) 2n

n 2

2 1

(B)

2n

m 1

2

(C)

2n 1

2n 2

2

(D*)

2n

m 2

2

Sol. Player should get (HT, HT, HT, ........) or (TH, TH, ......) atleast 2n times. If the sequence start from first

place then probability 2n�1

12

and if any other time probability is 2n

m

2.

Hence probability = 2n

m 2

2

[Type text]

3. Triplet (x, y, z) is chosen from the set {1,2,3,...,10}, such that x y < z. The number of such triplets is �

leqPp; {1,2,3,...,10} esa ls ,d f=kd (x, y, z) bl izdkj pquk tkrk gS fd x y < z ,sls f=kdks dh la[;k gS�

(A) 1000 (B) 120 (C) 45 (D*) 165 [PR-CD] [303] Sol. Any three numbers x, y, z from {1, 2, 3, .......} can be chosen in nC3 ways and we get unique triplet (x, y, z),

x < y < z. Again any two numbers x, z can be chosen from {1, 2, 3, ....,n} in nC2 ways and we get the triplet

(x, x, z), x < z. Hence total number of required triplets is nC2 + nC3.

Hindi leqPp; {1, 2, 3, .......} esa ls rhu la[;k,¡ x, y, z ds p;u ds rjhds nC3 rFkk gesa vf}rh; f=kd (x, y, z), x < y < z

izkIr gksrk gSA iqu% {1, 2, 3, ....,n} esa ls nks la[;k,¡ x ,oa z ds p;u ds rjhds nC2 bl izdkj izkIr f=kd (x, x, z), x < z

gSA vr% vHkh"B f=kdks dh la[;k nC2 + nC3.

4. If lines ax + by = 1, cx + dy = 1 and ex + fy = 1 are concurrent lines where a,b,c,d,e,f are nonzero different

real numbers then number of circles passes through (a,b), (c, d) and (e,f) are [DT-ES] [303] (A*) 0 (B) 1 (C) 2 (D) infinite

;fn js[kk,¡ ax + by = 1, cx + dy = 1 ,oa ex + fy = 1 laxkeh js[kk,¡ gks tcfd a,b,c,d,e,f fHkUu-fHkUu v'kwU; okLrfod

la[;k,¡ gS rc fcUnqvksa (a,b), (c, d) ,oa (e,f) ls xqtjus okys oÙ̀kksa dh la[;k gksxh-

(A*) 0 (B) 1 (C) 2 (D) vuUr

Sol. Lines ax + by = 1 cx + dy = 1 ex + fy = 1 are concurrent

a b 1

c d 1

e f 1

= 0

(a,b), (c,d), (e,f) are collinear

Number of circles passing through (a,b), (c,d), (e,f) are = 0

Hindi js[kk,¡ ax + by = 1

cx + dy = 1

ex + fy = 1 laxkeh gSA

a b 1

c d 1

e f 1

= 0

(a,b), (c,d), (e,f) lajs[kh; gSA

fcUnqvksa (a,b), (c,d), (e,f) ls xqtjus okys oÙ̀kksa dh la[;k = 0

5. If circle S = 0 passing through point of intersection of line 2x � 3y = 0 and circle x2 + y2 + 2x � 4y � 4 = 0

and distance of point (�5, 8) from centre of circle S = 0 is minimum then radius of circle is [CR-FC] [302] ;fn oÙ̀k S = 0 js[kk 2x � 3y = 0 ,oa oÙ̀k x2 + y2 + 2x � 4y � 4 = 0 ds izfrPNsnu fcUnqvksa ls xqtjrk gS rFkk fcUnq

(�5, 8) ls oÙ̀k S = 0 ds dsUnz dh nwjh U;wure gS rc oÙ̀k dh f=kT;k gS - (A) 31 (B*) 93 (C) 71 (D) 33 Sol. Equation of circle is (x2 + y2 + 2x � 4y � 4) + (2x � 3y) = 0 x2 + y2 + 2(1 + )x � (4 + 3)y � 4 = 0

centre C = 4 3

�(1 ),2

, radius = 2

2 4 3(1 ) 4

2

P(�5, 8) then CP = 213� 26 52

4

[Type text]

for CPmin. = � (�26)

132.

4

= 4 Radius = 93

Hindi oÙ̀k d k lehdj.k (x2 + y2 + 2x � 4y � 4) + (2x � 3y) = 0 x2 + y2 + 2(1 + )x � (4 + 3)y � 4 = 0

dsUnz C = 4 3

�(1 ),2

, f=kT;k = 2

2 4 3(1 ) 4

2

P(�5, 8) rc CP = 213� 26 52

4

CPmin ds fy;s. = � (�26)

132.

4

= 4 f=kT;k = 93

6. In the triangle ABC cosA + 2cosB = 2 � cosC then sides a,b,c are in [ST-HA] [303] (A*) A.P. (B) G.P. (C) H.P. (D) A.G.P. f=kHkqt ABC esa cosA + 2cosB = 2 � cosC rc Hkqtk,¡ a,b,c gksxh - (A*) lekUrj Js<h esa (B) xq.kksÙkj Js<h esa (C) gjkRed Js<h esa (D) lekUrj xq.kksÙkj Js<h esa Sol. cosA + 2cosB = 2 � cosC cosA + cosC = 2(1 � cosB)

2cosA C

2

cosA � C

2

= 4sin2 B2

cosA C

2

cosA � C

2

= 2sin2 A C

�2 2

cosA � C

2

= 2cosA C

2

cosA2

cosC2

+ sinA2

sinC2

= 2cosA2

cosC2

� 2sinA2

sinC2

cot. A2

cotC2

= 3

s(s � a)

(s � b)(s � c).

s(s � c)

(s � a)(s � b)= 3

2s = 3b a + b + c = 3b a,b,c are in A.P. l-Js- esa gksxs 7. If z1, z2,, z3 are the vertex of triangle such that |z1 � i| = |z2 � i| = |z3 � i| = 2 and z1 + z2 = 3i � z3 then area

of triangle is [CN-MD] [301]

;fn fdlh f=kHkqt ds 'kh"kZ z1, z2 ,oa z3 bl izdkj gS fd |z1 � i| = |z2 � i| = |z3 � i| = 2 rFkk z1 + z2 = 3i � z3 rc f=kHkqt

dk {ks=kQy gksxk -

(A) 6 3 (B*) 3 3 (C) 9 3 (D) 2 3

Sol. |z1 � i| = |z2 � i| = |z3 � i| i is circumcentre (i ifjdsUnz gS) ..........(1)

1 2 3z z z

3

= i i is centroid (i dsUnzd gSA) ..........(2)

(1) and (2) triangle is equilateral.

(1) rFkk (2) ls f=kHkqt leckgq gksxkA

[Type text]

60º

(

60º

Mz2 z3

z1

2

Mz3 = 2sin60º = 3

side (Hkqtk) (z2z3) = 2 3

Area of = 3

4 (2 3 )2 = 3 3

dk {ks=kQy = 3

4 (2 3 )2 = 3 3

8. If the function f(x) = x2 + bx + 3 is not injective for values of x in the interval 0 x 1, then complete set of

values of b will be [FN-RG] [301]

vUrjky 0 x 1, esa x ds ekuksa ds fy;s ;fn Qyu f(x) = x2 + bx + 3 ,dSdh ugh gS rks b ds ekuksa dk iw.kZ leqPp;

gksxk -

(A) (� , ) (B) (2, ) (C*) (�2, 0) (D) (�, �2) Sol. f(x) = x2 + bx + 3 Vertex lies in (0, 1)

'kh"kZ (0, 1) esa fLFkr gSA

0 < �b2

< 1

� 2 < b < 0

9. If f(x) be a twice differentiable function from R R such that t2f(x) � 2t f '(x) + f "(x) = 0 has two equal

values of t for all x and f(0) = 1, f '(0) = 2, then x 0lim

f(x) �1 t�

x 2

is [DE-MS] [304]

R R esa ifjHkkf"kr nks ckj vodyuh; Qyu f(x) bl izdkj gS fd t2f(x) � 2t f '(x) + f "(x) = 0 ls x ds lHkh ekuksa ds

fy;s t ds nks leku eku izkIr gksrs gS rFkk f(0) = 1 ,oa f '(0) = 2 rc x 0lim

f(x) �1 t�

x 2

gS -

(A) � 1 (B*) 1 (C) 12

(D) 0

Sol. t2f(x) � 2t f '(x) + f "(x) = 0 has equal roots (ds ewy ,d leku gS)

D = 0 f "(x)f '(x)

= f '(x)f(x)

ln (f (x)) = ln(f(x)) � n c

f(x) = cf(x) f(0) = cf(0) c = 12

f '(x)f(x)

= 2 ln(f(x)) = 2x + k ln (f(0)) = k k = 0

ln (f(x)) = 2x f(x) = e2x t2 .e2x � 4te2x + 4e2x = 0

t = 2

x 0lim

f(x) �1 t�

x 2

= x 0lim

2xe �1 22 �

2x 2

= 2 � 1 = 1

[Type text]

10. If the value of definite integral 1

�1 �1

2 2 |x|�1

1 xcot cot dx

1� x 1� (x )

= 2( a � b)

c

when a,b,c N in their

lowest form then a + b + c is equal to [DI-II] [302]

;fn 1

�1 �1

2 2 |x|�1

1 xcot cot dx

1� x 1� (x )

= 2( a � b)

c

ljyre :i esa gS tcfd a,b,c N rc

a + b + c d k eku gS -

(A*) 7 (B) 5 (C) 9 (D) 14

Sol. =1

�1 �1

2 2 |x|�1

1 xcot cot dx

1� x 1� (x )

= 1

�1 �1

2 2 |x|�1

1 �xcot cot dx

1� x 1� (x )

2 = 1

�1 �1 �1

2 2 |x| 2 |x|�1

1 x xcot cot � cot dx

1� x 1� (x ) 1� (x )

2 = 1

�1

2�1

1cot dx

1� x

2 = 1

�1 2

�1

tan 1� x dx

Integrating by parts [k.M'k% lekdyu djus ij

= 2

11�1 2

2 200

x �xx tan 1� x � dx

(1 1� x ) 1� x

= 2

1 2

2 20

x dx

(2 � x ) 1� x put x = sin

= 2

/2 2

20

sind

2 � sin

11. The equation of the curve satisfying the differential equation 2

2

d y

dx(x2 + 1) = 2x

dydx

passing through the point

(0, 1) and having slope of tangent at x = 0 is 3, is [DE-HO] [304]

vody lehdj.k 2

2

d y

dx(x2 + 1) = 2x

dydx

dks larq"V djus okys oØ dk lehdj.k] tks (0, 1) ls xqtjrk gS rFkk x = 0 ij

Li'kZ js[kk dh izo.krk 3 gS, gksxk&

(A) y = x2 + 3x + 2 (B) y2 = x2 + 3x + 1 (C*) y = x3 + 3x + 1 (D) y = x3 � 3x + 1

Sol. 2

2

d y

dx (x2 + 1) = 2x

dydx

2

2

d y

dxdydx

dx = 2

2x

x 1 dx

n dydx

= n (x2 + 1)+n c

[Type text]

dydx

= c(x2 + 1) c = 3 as at pawfd x = 0 ij, dydx

= 3

dydx

= 3(x2 + 1)

y = x3 + 3x + 1

12. If 1 2 3A ,A , A

and 1 2 3B ,B , B

are two sets of non-coplanar vectors such that r sA B

= 0 , if r s

5 , if r s

for

r = 1,2,3 and s = 1,2,3 then 1 2 3[ A A A ]

1 2 3[B B B ]

equals. [VT-ST] [307]

;fn 1 2 3A ,A , A

vkSj 1 2 3B ,B , B

vleryh; lfn'kksa ds nks leqPp; bl izdkj gS fd r sA B

= 0 , r s

5 , r s

; fn; fn

r =

1,2,3 ds fy, rFkk s = 1,2,3 rc 1 2 3[ A A A ]

1 2 3[B B B ]

cjkcj gS -

(A) 0 (B) 25 (C*) 125 (D) 250

Sol. Let ekuk 1A

= 1 2 3� � �i j k ; 2A

= 1 2 3� � �i j k ; 3A

= 1 2 3� � �i j k

1B

= 1 2 3� � �x i x j x k ; 2B

= 1 2 3� � �y i y j y k ; 3B

= 1 2 3� � �z i z j z k

1 2 3[ A A A ]

1 2 3[B B B ]

= 1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

x x x

y y y

z z z

= 1 1 2 2 3 3 1 1 2 2 3 3 1 1 2 2 3 3

1 1 2 2 3 3 1 1 2 2 3 3 1 1 2 2 3 3

1 1 2 2 3 3 1 1 2 2 3 3 1 1 2 2 3 3

x x x y y y z z z

x x x y y y z z z

x x x y y y z z z

= 1 1 1 2 1 3

2 1 2 2 2 3

3 1 3 2 3

A .B A .B A .B

A .B A .B A .B

A .B A .B A .B3

=

5 0 0

0 5 0

0 0 5

= 125

Comprehension (13-14) If a die is thrown thrice and its number is show by the coefficient of the quadratic equation ax2 + bx + c = 0

as a, b and c. Then answer the following [PR-CD] [301]

;fn ,d iklk rhu ckj Qsdk tkrk gS rFkk bldh la[;k f}?kkr lehdj.k ax2 + bx + c = 0 ds xq.kkdksa a, b vkSj c ls

n'kkZ;h tkrh gSA rc fuEu iz'uksa dk mÙkj nhft,A 13. The probability that roots are real, is

izkf;drk fd ewy okLrfod gS -

[Type text]

(A) 38

216 (B*)

43216

(C) 5

216 (D)

6216

14. If a, b, c are in A.P., then the probability that roots are real is

;fn a, b, c lekUrj Js.kh esa gS rc izkf;drk tcfd ewy okLrfod gS] gksxh-

(A) 1

216 (B)

2216

(C) 3

216 (D*) 0

Sol. For real roots okLrfod ewyksa ds fy, D 0 b2 � 4ac 0

b (a, c) Fav. Case

2 (1, 1) 1

3 (1, 1), (1, 2), (2, 1) 3

4 (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (3, 1) 8

5(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (5, 1), (6, 1)

14

6(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1), (6, 1)

17

13. Required probability = 43

216

vHkh"V izkf;drk = 43

216

14. There is no such case for which b2 � 4ac 0 and 2b = a + c

required probability = 0

216 = 0

bl izdkj dh dksbZ fLFkfr ugh gS fd b2 � 4ac 0 vkSj 2b = a + c

vHkh"V izkf;drk = 0

216 = 0

Comprehension # 2 If P(x, y) means to make the point (x, y) on x-y plane dark [DI-LS] [304]

;fn P(x, y) dk vFkZ gS fd fcUnq (x, y) lery x-y ij xgjs dkys jax ls n'kkZ;k gSA

P(x1, y1) + P(x2, y2) means to make both points (x1, y1) and (x2, y2) as dark point on x-y plane.

P(x1, y1) + P(x2, y2) dk vFkZ gS fd nksuksa fcUnqvksa (x1, y1) rFkk (x2, y2) lery x-y ij xgjs dkys jax ls n'kkZ;k gSA

Now answer the following problems:

vc fuEu iz'uksa dk mÙkj nhft,A

15. nlim

2n

r 1

rP 1,

n

represents a

[Type text]

(A) triangle (B) circle (C*) line segment (D) square

nlim

2n

r 1

rP 1,

n

O;Dr djrk gS -

(A) f=kHkqt (B) oÙ̀k (C*) js[kk[k.M (D) oxZ

16. nlim

22n

r 1

r rP ,

n n

represents a

(A*) parabolic segment (B) elliptical segment (C) circle (D) arc of circle

nlim

22n

r 1

r rP ,

n n

O;Dr djrk gS -

(A*) ijofyd [k.M (B) nh?kZoÙ̀kkdkj [k.M

(C) oÙ̀k (D) oÙ̀k d k pki

Paragraph for Question Nos. 17 to 18

ABC is a triangle, O is a point inside the triangle so that its distance from A,B,C is respectively a,b,c. L,M,N are the feet of the perpendiculars from O to AB, BC, CA respectively x, y, z are respectively the distances

of O from L,M,N. OAL = , OBM = , OCN = [TR-MS] [304] [Old, E] 17. AL + BM + CN is equal to

(A*) a cos + b cos + c cos (B) a sin + b sin + c sin

(C) x cos + y cos + z cos (D) x sin + y sin + z sin

18. sin (A � ) + sin (B � ) + sin (C � ) is equal to

(A) x y za b c (B*)

x y zb c a (C)

x y zc a b (D) x cot + y cot+ z cot

Sol. (17, 18)

AL = a cos = x cot Al + BM + CN = a cos + b cos + c cos

BM = b cos = y cot

CN = c cos = z cot

sin (A � ) = za

, sin (B � ) = xb

, sin (C � ) = yc

sin (A � ) + sin (B � ) + sin (C � ) = x y zb c a

a = x cosec etc.

x cosec + y cosec + z cosec = a + b + c

ç'u 17 ls 18 ds fy, vuqPNsn

[Type text]

ABC ,d f=kHkqt gS] f=kHkqt ds vUnj ,d fcUnq O bl çdkj gS fd bldh A,B,C ls nqfj;ka Øe'k% a,b,c gSA rFkk O ls

Hkqtk AB, BC, CA ij yEcikn Øe'k% L,M,N gS vkSj O ls L,M,N dh nqfj;ka Øe'k% x, y, z gSA

OAL = , OBM = , OCN = [Old, E]17. AL + BM + CN =

(A*) a cos + b cos + c cos (B) a sin + b sin + c sin

(C) x cos + y cos + z cos (D) x sin + y sin + z sin

18. sin (A � ) + sin (B � ) + sin (C � ) =

(A) x y za b c (B*)

x y zb c a (C)

x y zc a b (D) x cot + y cot+ z cot

Sol. (17, 18)

AL = a cos = x cot Al + BM + CN = a cos + b cos + c cos

BM = b cos = y cot

CN = c cos = z cot

sin (A � ) = za

, sin (B � ) =xb

, sin (C � ) = yc

sin (A � ) + sin (B � ) + sin (C � ) = x y zb c a

a = x cosec etc.

x cosec + y cosec + z cosec = a + b + c

Paragraph for Question Nos. 19 to 20 Let a,b,c are the roots of the equation x3 � 3x2 + 5x � 1 = 0. A and B are two matrices given as

A =

bac

acb

cba

, B =

222

222

222

b�caa�bcc�ab

a�bcc�abb�ca

c�abb�caa�bc

[MT-AL] [303]

If , and are the roots of the equation |B � x| = 0, where is the unit matrix of order 3.

ekuk a, b, c lehdj.k x3 � 3x2 + 5x � 1 = 0 ds ewy gSA A vkSj B nks vkO;wg bl izdkj gS fd

A =

bac

acb

cba

, B =

222

222

222

b�caa�bcc�ab

a�bcc�abb�ca

c�abb�caa�bc

;fn , rFkk lehdj.k |B � x| = 0 ds ewy gS tgk¡ 3 Øe dk bdkbZ vkO;wg gS 19. The value of the det (A). det (B) is equal to

det (A). det (B) dk eku cjkcj gS -

(A) 18 (B) 182 (C*) 183 (D) 184

20. The value of + + is equal to + + dk eku cjkcj gS - (A) 3 (B*) 6 (C) 9 (D) 18

[Type text]

Sol. B is the cofactor matrix of A B, A dk lg[k.M vkO;wg gSA |B| = |A|2

19. det (A.B) = |A| |B| = |A|3 |A| = 3abc �a3 � b3 � c3 = � (a + b + c) (a2 + b2 + c2 � ab � bc � ca) =�(a + b + c) [(a + b + c)2 � 3(ab + bc + ca)] = � 3[9 � 3 × 5] = 18 |AB| = |A|3. Ans

20. |B � x| = 0

+ + = tr |B| = ab + bc + ca � (a2 + b2 + c2) = 3(ab + bc + ca) � (a + b + c)2 = 3 × 5 � 9 = 6 Integer Type (two digits)

21. Let A(0, 1), B and C are three points on 2

2xy 1

4 . Equation of altitude from A on BC is x = 0. Length of

side of this equilateral triangle is , then 13

3

is equal to. [EL-MS] [302]

ekuk A(0, 1), B vkSj C oØ 2

2xy 1

4 ij rhu fcUnq gSA A ls BC ij 'kh"kZ yEc dk lehdj.k x = 0 gS bl leckgq

f=kHkqt ds Hkqtk dh yEckbZ gS rc 13

3

cjkcj gS&

Ans. 16 Sol. Equation of ellipse is

2x

4 +

2y1

= 1 ...(i)

Given that equation of altitude is x = 0 from A that means one side parallel to x-axis and given that length of the side is so

(2,0)

(0,1)

x

y

B

60º

30º

(�2,0)

(0,�1)

A

C

triangle is equilateral.

Equation of AB will be y = 3x + 1 ...(ii) Solving (i) and (ii)

x = 0, � 8 313

y = � 2413

+ 1 = � 1113

, 1

Coordinate of B = 8 3 11

� ,�13 13

So, = 64 3 24 24169 169

= 16 3

13 13 =16 3 .

Hindi nh?kZoÙ̀k dk lehdj.k gS

[Type text]

2x

4 +

2y1

= 1 ...(i)

fn;k x;k gS fd A ls 'kh"kZ yEc dk lehdj.k x = 0 gS] dk vFkZ ,d Hkqtk x-v{k ds lekUrj gS rFkk nh xbZ Hkqtk dh

yEckbZ gS blfy,

(2,0)

(0,1)

x

y

B

60º

30º

(�2,0)

(0,�1)

A

C

f=kHkqt ,d leckgq f=kHkqt gSA

AB dk lehdj.k gksxk y = 3x + 1 ...(ii)

(i) vkSj (ii) gy djus ij

x = 0, � 8 313

y = � 2413

+ 1 = � 1113

, 1

B ds funsZ'kkad = 8 3 11

� ,�13 13

blfy, , = 64 3 24 24169 169

= 16 3

13 13 =16 3 .

22. Let the function g(x) = f(x2 � x � 10) + f(14 + x � x2), f(x) > 0 for all real numbers x except finite number of

real numbers x1 for which f(x) = 0. Then find the number of negative integral values of x for which g(x) is

increasing. [MN-MI] [303]

ekuk Qyu g(x) = f(x2 � x � 10) + f(14 + x � x2), okLrfod la[;kvksa dh fuf'pr la[;k x1 ftuds fy,

f(x) = 0 gS] dks NksMdj lHkh okLrfod la[;kvksa x ds fy, f(x) > 0 gS] rc g(x) ds o/kZeku gksus ds fy, x ds

_ .kkRed iw.kkZad ekuksa dh la[;k Kkr dhft,A Ans. 03

Sol. f(x) > 0 f(x) is increasing f(x) o/kZeku gSA

Now g(x) = (2x � 1) [f(x2 � x � 10) � f (14 + x � x2) 0

Case fLFkfr -1 : x 12

and vkSj x 4 x [4, )

Case fLFkfr -2 2x � 1 0 and vkSj f (x2 � x � 10) f (14 + x � x2)

x 12

and vkSj � 3 x 4 x 1

3,2

23. Find the number of 3 digit natural numbers whose consecutive digit are in geometric progression

3 vad dh izkd r̀ la[;kvksa dh la[;k Kkr dhft, ftlds Øekxr vad xq.kksÙkj Js.kh esa gSA [PC-MS] [304]

Ans. 17 Sol. If common ratio is 1 numbers are 111,222,...... , 999

[Type text]

If common ratio is 2( or 12

) numbers are 124, 248 (421, 842)

If common ratio is 3( or 13

) numbers are 139 (931)

If common ratio is 32

( or 23

) numbers are 469 (964)

Hence total = 17

Hindi ;fn lkoZ vuqikr 1 gS la[;k,¡ 111,222,...... , 999 gS

;fn lkoZ vuqikr 2(;k 12

) gS rc la[;k,¡ 124, 248 (421, 842) gS

;fn lkoZ vuqikr 3(;k 13

) gS rks la[;k,¡ 139 (931) gS

;fn lkoZ vuqikr 32

(;k 23

) la[;k,¡ 469 (964) gS

vr% dqy = 17

24. A circle in a plane divides the plane in two parts. Find how many maximum parts can 10 circles divides the

plane. [CR-MS] [304]

,d lery esa ,d oÙ̀k nks Hkkxksa esa lery dks foHkkftr djrk gSA 10 oÙ̀k] lery dks fdrus vf/kdre Hkkxksa esa foHkDr

djrk gSA Ans. 92 Sol. Total number of parts = number of point of intersection + 2

Hkkxksa dh dqy la[;k = izfrPNsn fcUnqvksa dh la[;k + 2

= 2 . nC2 + 2 f(x) = n2 � n + 2 f(10) = 92

25. A,B,C, D are four points in the space and satisfy AB

= 3, BC

= 7, CD

= 11 and DA

= 9, then how

many values are possible of AC BD

? [VT-AD] [302]

lef"V esa pkj fcUnq A, B, C, D gS rFkk AB

= 3, BC

= 7, CD

= 11 vkSj DA

= 9 gS, rc AC BD

ds laHkkfor eku

fdrus gS ?

Ans. 01

Sol. 2 2| AB | | CD |

= 32 + 112 = 130 = 72 + 92 = 2 2| BC | | DA |

AB BC CD DA 0

(As it will form a closed loop) (pwfd ;g cUn ywi ds :i dk gksxkA)

2 2| (DA) | | (AB BC (CD) |

2 2 2 2| DA | | AB | | BC | | CD | 2(AB.BC BC.CD CD.AB)

22 2 2 2| DA | | AB | � | BC | | CD | 2(BC AB.BC BC.CD CD.AB)

2 2 2 2| DA | | BC | | AB | | CD | 2[(AB BC).(BC CD)]

AC.BD

= 0

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Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005

Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029

P1-1

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1 to 12 SCQ 12 3 �1 36



26 to 37 SCQ 12 3 0 36



51 to 62 SCQ 12 3 0 36



75 225

Paper-1

Total Total

Maths

Physics

Chemistry

PAPER-1

SECTION-1 : (Only One option correct type) [k.M�1 : (dsoy ,d lgh fodYi çdkj)

This section contains 12 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

bl [k.M esa 12 cgqfodYi ç'u gSA çR;sd ç'u esa pkj fodYi (A), (B), (C) vkSj (D) gS] ftuesa ls dsoy ,d lgh gSA

SCQ._(12)

26. In the given circuit potential difference across the capacitor in steady state is

fn;s x;s ifjiFk esa LFkk;h voLFkk esa la/kkfj=k ds fljksa ij foHkokUrj gksxk % [CP-CK](104)

(A) 1 V (B) 5

V2

(C*) 6V (D) 3V

Sol. In steady state current through capacitor will be zero.

LFkk;h voLFkk esa la/kkfj=k ls xqtjus okyh /kkjk 'kwU; gksxh

So equivalent circuit is given below.

http://www.resonance.ac.in

mailto:[email protected]

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P1-2

vr% rqY; ifjiFk uhps fn;k x;k gSA

10 V 2V8V

2A 1A

10 V

0 V 27. A uniform ring of mass 1 kg and radius 1 m rolls down an inclined plane of inclination 30° starting

from rest without slipping. At the instant when the centre of the ring has travelled a distance 20 m along the incline plane, choose the correct option : [RB-CD](104)

1 kg nzO;eku rFkk 1 m f=kT;k dh ,dleku oy;] 30° >qdko ds ur ry ij fojke ls fcuk fQlys uhps dh

vksj yq<+drh gSA tc oy; dk dsUnz ur ry ds vuqfn'k 20 m nwjh r; djrk gS] rks bl {k.k ds laxr lgh

dFku NkafV;s %

(A) Velocity of centre of mass is 20 m/s

(B) Angular velocity of ring is 5 rad./s

(C) Angular velocity of ring is 20 rad./s

(D*) Angular momentum about its centre of mass 10 kg m2/s

(A) nzO;eku dsUnz dk osx 20 m/s gSA

(B) oy; dk dks.kh; osx 5 rad/s gSA

(C) oy; dk dks.kh; osx 20 rad/s gSA

(D*) blds nzO;eku dsUnz ds ifjr% dks.kh; laosx 10 kg m2/s gSA

Sol. Ring perform pure rotation about point of contact so

oy; lEidZ fcUnq ds lkis{k 'kq} ?kw.kZu djrh gS vr%

mg sin = 2 21(2mR )

2 =

g sin

R

Alternative Sol.

Let the velocity of centre of mass of ring vcm and its angular speed be �� after rolling a distance

on inclied plane.



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P1-3

vcm = R ....(1)

From conservation of energy

mg sin = 12

cm 2 + 12

m vcm2 =

12

(mR2)2 + 12

m (R22) = mR22

= g sin

R

Hence the angular momentum of ring at the lowest point about an axis passing through centre of mass is

s Lcm = cm = mR2 = mR g sin = 10 kg m2/s

gy% ekuk oy; ds nzO;eku dsUnz dk osx vcm rFkk bldh dks.kh; pky �� gksxh] tc ;g ur ry ij yksVuh xfr dj

nwjh r; djrh gSA

vcm = R ....(1)

Å tkZ laj{k.k ls

mg sin = 12

cm 2 + 12

m vcm2 =

12

(mR2)2 + 12

m (R22) = mR22

= g sin

R

nzO;eku dsUnz ds ifjr% oy; dk dks.kh; laosx

Lcm = cm = mR2 = mR g sin = 10 kg m2/s

28. When a non-isotropic cube is heated its length expands with coefficient of linear expansion breadth expands with coefficient of linear expansion 2 and height expands with coefficient of linear expansion 3The coefficient of volume expansion will be : [CT-TE](104)

tc ,d vlenSf'kd ?ku dks xeZ fd;k tkrk gSA rc bldh yEckbZ js[kh; izlkj xq.kkad pkSM+kbZ js[kh; izlkj

xq.kkad 2 ,oa Å WpkbZ js[kh; izlkj xq.kkad 3 ls izlkfjr gksrh gS] rc vk;ru izlkj xq.kkad gksxk % (A) = (B) = 3 (C) = 2 (D*) = 6



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P1-4

Sol. V = bh

dV d db dhV b h

T = T + 2T + 3T

= 6 29. A hollow thick spherical shell of inner radius a and outer radius b carries volume charge density

2

kr

where k is some positive constant and r is distance from centre. Select correct variation of

magnitude of electric field with distance r. (b 2a) [ES_EF] (108)

a vkarfjd f=kT;k rFkk b ckg~; f=kT;k ds ,d [kks[kys eksVs xksyh; dks'k dk vk;ruh; vkos'k ?kuRo 2

k

r gSA

tgk¡ k dqN /kukRed fu;rkad gS rFkk r dsUnz ls nwjh gSA fo|qr {kS=k ds ifjek.k dk nwjh r ds lkFk ifjorZu dk

lgh xzkQ gksxk (b 2a)

ab

(A)

E

a b r

K(b - a)b2

(B)

E

r

K(b - a)b2

(C*)

E

a b r

K(b - a)b2

(D)

E

r

Sol. for r < a ds fy,

since pwafd Qin = 0

so vr% E = 0 ......(1)

for a r b ds fy,

Qenclosed = r

22

a

k.4 r dr

r

Qenclosed r

a

4 k dr

Qenclosed = 4k(r � a)

E = enclosed2

0

Q14 r



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P1-5

so vr%,

2 20 0

4 k r a k r aE

4 r r

.......(2)

Now vc, r b

Qenclosed = b

22

a

k.4 r dr 4 k b a

r

E = enclosed2

0

Q14 r

= 2

K(b a)

r

30. The figure shows a conical container of half-apex angle 37o filled with certain amount of kerosene and water as shown. Specific gravity of kerosene is 0.8 and axis of cone is vertical. The force exerted by the water on the kerosene is approximately.

(Take atmospheric pressure = 105 Pa and g = 10m/s2). [FL-PR](102)

37° v)Z 'kh"kZ dks.k okyk 'kadqokdkj ik=k fp=kkuqlkj dqN ek=kk esa dsjksflu rFkk ikuh ls Hkjk gqvk gSA dsjksflu

dk fof'k"V xq:Ro 0.8 rFkk 'kadq dh v{k Å /okZ/kj gSA ikuh }kjk dsjksflu ij vkjksfir cy yxHkx gksxkA

(ok;qe.Myh; nkc = 105 Pa rFkk g = 10m/s2).

10 m

8 m

P = 10 Pa05

(A) 3 × 106 N (B) 4 × 107 N (C*) 2 × 107 N (D) 5 × 107 N

Sol. Density of kerosene is 800 kg/m3 and density of water is 1000 kg/m3.

dSjkslhu dk ?kuRo 800 kg/m3 rFkk ikuh dk ?kuRo 1000 kg/m3 gSA

r

tan378

= 34

�

r

37°

r = 6m

F = (P0 + hg) r2

= (105 + 10 × 800 × 10) × × 36



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P1-6

1.8 × 36 × × 105

= 2 × 107 N

31. A soap bubble is being blown slowly on a tube of radius 1 cm. The surface tension of the soap solution is 0.05 N/m and surface of the bubble makes an angle of 60o with the tube as shown. The excess pressure over the atmospheric pressure in the tube is

,d lkcqu ds cqycqys dks 1 cm f=kT;k okyh uyh ij /khjs&/khjs Qqyk;k tkrk gSA lkcqu ds foy;u dk i"̀B ruko 0.05 N/m rFkk cqycqys dh lrg }kjk uyh ls cuk;k x;k dks.k fp=kkuqlkj 60° gks rks ok;qe.Myh; nkc dh rqyuk esa uyh esa nkc vkf/kD; gksxk : [SF-EX](104)

(A) 5 Pa (B) 1 Pa (C*) 10 Pa (D) 32 Pa

Sol.

R = 1cm

cos60 = 2cm

P = 4SR

= 2

4 0.0510 Pa

2 10

32. A plane wave front of light, whose wavelength can be varied continuosly, falls normally on a uniform thin film of oil which covers a glass plate. Constructive interference is observed in reflected

light for = 5000 Å and

= 7500 Å and for no other wavelength in between. If of oil is 1.25

and that of glass is 1.5, the thickness of oil film nearly will be : [YE-TF](104)

izdk'k dk ,d lery rjaxzkx] ftldh rjaxnS/;Z yxkrkj ifjofrZr gks ldrh gS] rsy dh iryh le:i ijr ij

yEcor~ vkifrr gS] tks dk¡p d h IysV ij iw.kZr% QSyh gqbZ gSA izdk'k ds = 5000 Å rFkk

= 7500 Å ds

fy, ijkofrZr izdk'k esa lEiks"kh O;frdj.k izsf{kr gksrk gS rFkk bu nksuksa rjaxnsZ/;ksa ds e/; fdlh Hkh vU;

rjaxnsZ/;Z ds fy, lEiks"kh O;frdj.k izsf{kr ugha gksrk gSA ;fn rsy dk viorZukad = 1.25 rFkk dkWp dk

viorZukad 1.5 gks rks rsy dh ijr dh eksVkbZ yxHkx gksxhA

(A) 15

µm (B) 23

µm (C*) 35

µm (D) 23

µm

Sol. 20t = n × 7500 = (n + 1) × 5000

n = 2 t =35

µm



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P1-7

33. The shape of a wave pulse in a string propagating in either the positive x-direction or in negative x-

direction is given y = 2

1

1 xat t = 0s and y =

2

1

2 2x x

at t = 1s respectively, where x and y

are in meters. The shape the wave disturbance does not change during propagation. The velocity of the wave is : [ST-EQ](104)

jLlh ij /kukRed x- fn'kk ;k _ .kkRed x-fn'kk esa lapfjr rjax LiUn dh vkd f̀Ùk t = 0 lSd.M ij

y = 2

1

1 x rFkk t = 1 lSd.M ij y =

2

1

2 2x x

}kjk nh tkrh gS] ;gkW x rFkk y ehVj es gSA lapj.k ds

nkSjku rjax fo{kksHk dh vkd f̀Ùk ifjofrZr ugha gksrh gSA rjax dk osx gksxkA

(A*) 1 m/s in positive x direction (B) 2 m/s in negative x direction

(C) 12

m/s in positive x direction (D) 12

m/s in negative x direction

(A*) /kukRed x- fn'kk esa 1 m/s (B) _ .kkRed x- fn'kk esa 2 m/s

(C) /kukRed x- fn'kk esa 12

m/s (D) _ .kkRed x- fn'kk esa 12

m/s

Sol. y = f(x ± c · t) is the general wave equation

y = f(x ± c · t) ,d O;kid rjax lehdj.k gSA

At t = 0 ij, y = f(x) y = 2

1

1 x

y = 2

1

2 2x x =

2

1

1 (x 1) = f (x � 1)

f (x � ct) = f (x � 1) at t = 1 ij

c = 1 m/s. ]

34. If the surface of a metal is exposed, by two different radiations, of wavelength 1 = 350 nm and 2

= 450 nm. Maximum velocity of photoelectrons are obseved to be in the ratio of 2 : 1. Work function of the metal is : [MP-PE](102)

;fn ,d /kkrq dh lrg ij 1 = 350 nm rFkk 2 = 450 nm rjaxnS/;Z okyh nks fHkUu&fHkUu fofdj.k vkifrr gSA

rc QksVksbySDVªkWu ds vf/kdre~ osx dk vuqikr 2 : 1 izsf{kr gksrk gSA /kkrq dk dk;ZQyu gksxkA

(A) 2.84 × 10� 18 J (B) 1.6 × 1019 J

(C*) 4 × 10�19 J (D) 2.4 × 10�17 J



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P1-8

Sol. Let be the work function of metal.

ekuk /kkrq dk dk;ZQyu gSA

for first case : 1

hc

= + 21mv

2

izFke fLFkfr easa 1

hc

= + 21mv

2

for second case f}rh; fLFkfr esa : 2

hc

= + 22mv

2

hc 1 2

1 1

= 2 2

1 2m

v v2

= 2 22 2

m4v v

2

=223mv

2

22mv

2 =

1 2

hc 1 13

= 22

2

mvhc2

=2 1 2

hc hc 1 13

=

2 1

4hc hc3 3

= � 1 2

hc 1 43

= 34 8

9 9

6.62 10 3 10 4 13 450 10 350 10

= 4 × 10�19 J.

35. A square loop of area 2.5 × 10�3 m2, having 100 turns and net resistance of 100 is moved out of a uniform magnetic field of 0.40 T perpendicular to the magnetic filed in 1 sec with a constant velocity as shown in the figure. Work done, in pulling the loop by the external agent is. [EI-CM](104)

,d oxkZdkj ywi ftldk {kS=kQy 2.5 × 10�3 m2 rFkk ?ksjks dh la[;k 100 rFkk dqy izfrjks/k 100 gS] bls

le:i pqEcdh; {kS=k 0.40 T ls 1 sec esa fu;r pky ls pqEcdh; {kS=k ds yEcor~ xfr djkrs gq;s fp=kkuqlkj

ckgj fudkyk tkrk gS] rks ywi dks [khapus esa ckg~; dkjd }kjk fd;k x;k dk;Z gksxkA[Made by ABJ Sir 2013-14]

(A*) 100 J (B) 50 J (C) 200 J (D) 500 J

Sol. Work done = Q (heat dissipated)



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P1-9

fd;k x;k dk;Z = Q (O;f;r Å "ek)

= 2e

tR

= 2(N B v)

R

× t = 2 2 3100 (0.4) 2.5 10

100

× v2 × 1

2 = 2.5 × 10�3

= 425 10 = 5 × 10�2 m

v = 25 10

1

Work done = Q (heat dissipated) = 6.25 × 10�6 × 16 J = 100 µJ

fd;k x;k dk;Z = Q (O;f;r Å "ek) = 6.25 × 10�6 × 16 J = 100 µJ

36. A sample of He gas is undergoes a cyclic process ABCDA as shown. Here symbols have their usual meaning. Then which of the following options is not true. [TH-FL](104)

ghfy;e xSl ds ,d izfrn'kZ dks pØh; izØe ABCDA ls fp=kkuqlkj xqtkjk tkrk gSA ;gkW izfrdksa dk lkekU;

vFkZ gS] rks fuEu esa ls dkSulk fodYi lgh ugha gSA

d sYohu esa

esa

(A) A

B

P

P= 1 (B) B

A

T

T = 2 (C) BC

DA

| Q |

| Q |

= 1 (D*) max

min

P

P = 3

Sol. Number of moles = n (assume)

CV = 32

R and CP = 52

R (Monoatomic)

TA = 300 K

TB / TA = 2/1

Hence TB = 600 K



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P1-10

QBC = WBC = nRTB In C

B

V

V

= (n) (R) (600) ln 0

0

4V

2V

= 600 n R ln 2

QDA = WDA = nRTD In A

D

V

V

= (n) (R) (300) In 0

0

V

4V

= 300 n R n14

QDA = � 600 nR ln 2

B C

D A

Q

Q

= 1

Pressure is constant from A to B.

Pressure decreases from B to C and again from C to D.

Pressure increases from D to A.

So max

min

P

P= A

D

P

P = A A D

D D A

P V VP V V

= A

D

nRT 4nRT 1

= 4

eksyks dh la[;k = n (ekuk)

CV = 32

R rFkk CP = 52

R (,d ijek.koh;)

TA = 300 K

TB / TA = 2/1

vr% TB = 600 K

QBC = WBC = nRTB In C

B

V

V

= (n) (R) (600) ln 0

0

4V

2V

= 600 n R ln 2

QDA = WDA = nRTD In A

D

V

V

= (n) (R) (300) In 0

0

V

4V

= 300 n R n14

QDA = � 600 nR ln 2

B C

D A

Q

Q

= 1

A ls B rd nkc fu;r gSA

B ls C rd nkc ?kVrk gS rFkk iqu% C ls D rd nkc ?kVrk gSA

D ls A rd nkc c<+rk gSA



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P1-11

vr% max

min

P

P= A

D

P

P = A A D

D D A

P V VP V V

= A

D

nRT 4nRT 1

= 4

37. A uniform cylinder of mass m and radius r is suspended from a massless string as shown. One end of the string is attached directly to a rigid support, while the other end is attached to an ideal spring of spring constant k. Time period of the small vibrations of the cylinder is : (assume no sliding anywhere) [SH-SM](102)

m nzO;eku rFkk r f=kT;k dk ,d le:i csyu nzO;ekughu jLlh }kjk fp=kkuqlkj yVdk gqvk gSA jLlh d k ,d

fljk lh/ks n<̀+ vk/kkj ls tqM+k gqvk gS tcfd nwljk fljk k cy fu;rkad dh vkn'kZ fLiazx ls tqM+k gqvk gSA csyu

ds vYi nksyuksa dk vkorZdky gksxkA (ekuk dgh Hkh fQlyu ugha gSA)

r

K

(A*) 3m2

8k (B) 8k

23m

(C) 8m2

3k (D) k

2m

Ans. T = 2 3m8k

Sol. E = � mgX + 21 1mv 1

2 2

+ 2

0K

( 2x)2

dEdX

= � mg + m dV 3 K

2V2 dX 2 2 2(0 + 2x)2 = 0

2

2

d X 8Kx

3mdt

Alternative Sol.

21k(2r ) �mgr

2 +

22 21 mr

mr C2 2

12

k4r2 23mr

2 è �mgr è4

2 = 0

8k

�3m

+ constant (fu;r)

T = 3m

28k



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P1-12

SECTION � 2 : (Paragraph Type) [k.M � 2 : (vuqPNsn çdkj)

This section contains 4 paragraphs each describing theory, experiment, data etc. Eight questions relate to four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct answer among the four choices (A), (B), (C) and (D).

bl [k.M esa fl)karksa] ç;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 4 vuqPNsn gSA pkjksa vuqPNsnksa ls lacaf/kr vkB

ç'u gSa] ftuesa ls gj vuqPNsn ij nks ç'u gSaA fdlh Hkh vuqPNsn esa gj ç'u ds pkj fodYi (A), (B), (C) vkSj

(D) gSa] ftuesa ls dsoy ,d gh lgh gSA


iz'u 38 ls 39 ds fy, vuqPNsn

Three balls A, B and C each of mass m and same size, are placed along same line on smooth horizontal surface. A is given a velocity u towards B as shown. Coefficient of restitution for collision

between A & B is 12

and between B & C is 13

. Whole situation is shown in figure. Answer the

following questions (Assume all the collisions to be headon)

rhu xsan A, B rFkk C ftuesa izR;sd dk nzO;eku m rFkk vkdkj leku gS, fpduh {kSfrt lrg ij ,d js[kk ds

vuqfn'k fLFkr gSA xsan A dks fp=kkuqlkj xsan B dh rjQ osx u fn;k t krk gSA ;fn A rFkk B ds e/; VDdj dk

izR;koLFkku xq.kkad 12

,oa B rFkk C ds e/; izR;koLFkku xq.kkad 13 gSA lEiw.kZ fLFkfr fp=kkuqlkj gSA fuEu iz'uksa

ds mÙkj nhft,A (lHkh VDdjksa dks lEeq[k ekusa) [CO-HC](104)

38. Final relative velocity of B with respect to A will be : [CO-HC](104)

B dk A ds lkis{k vfUre lkis{k osx gksxkA

(A) u4

(B*) zero (C) u2

(D) u3

(A) u4

(B*) 'kwU; (C) u2

(D) u3

39. Total energy loss due to all collision will be : [CO-HC](104)

lHkh VDd jksa ds dkj.k dqy Å tkZ gkfu gksxhA

(A) 3

16 mu2 (B)

716

mu2 (C*) 5

16 mu2 (D)

916

mu2

Sol. m mm mv1

u v2

Just before collision Just after collision



PHYSICS

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P1-13

applying conservation of linear momentum mu = mv2 + mv1

js[kh; laosx lj{k.k }kjk mu = mv2 + mv1

coefficient of restitution v2 � v1 = eu

izR;koLFkku xq.kkad v2 � v1 = eu

v2 = u2

(1 + e) v1 = u2

(1 � e)

for collision between A and B for collision between B and C

A rFkk B ds e/; VDdj ds fy, B rFkk C ds e/; VDdj ds fy,

vB = u 1

12 2

=

3u4

vC = 1 3u 1 u

12 4 3 2

vA = u 1

12 2

=

u4

vB = 1 3u 1 u

12 4 3 4

Total number of collisions = 2

dqy VDdjksa dh la[;k = 2

Total energy loss E = 2

21 m 1 1 m 3u 1u 1 1

2 2 4 2 2 4 9

dqy Å tkZ esa gkfu E = 2

21 m 1 1 m 3u 1u 1 1

2 2 4 2 2 4 9

= 2 21 3 1 9 8mu mu

4 4 4 16 9

= 2

23mu 1mu

16 8

= . 25mu

16


iz'u 40 ls 41 ds fy, vuqPNsn

Compression-2

Consider the given R-C circuit with an ideal voltmeter, while charging (i.e., K1 closed) at t = 0, initial

voltmeter reading is V0/4. When voltmeter reading approaches to 03V

4 during charging switch K1 is

opened and K2 is closed. The switch K2 is closed till the reading of voltmeter again drops to 0V

4.



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P1-14

This cycle is continuously repeated (i.e., charging from 0V

4 to 03V

4 and then discharging from

03V

4 to 0V

4). Answer the following questions for this setup.

[CP-EQ](105)

C

VR

K2

K1

V0

vkn'kZ oksYVehVj okys ,d R-C ifjiFk dh dYiuk dhft,A vkos'ku ds nkSjku (vFkkZr~ tc K1 cUn gS) t = 0 ij

oksYVehVj dk izkjfEHkd ikB~;kad V0/4 gSA vkos'ku ds nkSjku tc oksYVehVj dk ikB~;kad 03V

4 rd igqWprk gS

rc dqath K1 dks [kksy fn;k tkrk gS rFkk dqath K2 dks cUn dj fn;k tkrk gSA daqth K2 rc cUn jgrh gS tc

rd oksYVehVj dk ikB~;kad nqckjk ?kVdj 0V

4 gksrk gSA bl izfØ;k dks nqckjk nksgjk;k tkrk gSA (vFkkZRk~ 0V

4ls

03V

4 rd vkosf'kr fd;k tkrk gS rFkk fQj 03V

4 ls 0V

4 rd fujkosf'kr fd;k tkrk gSA) fuEu iz'uksa ds mÙkj

nhft,A

C

VR

K2

K1

V0

40. Time period of the cycle is : [CP-EQ](105)

bl izfØ;k dk vkorZ dky gksxk %

(A) RC n 3 (B) RC n 2 (C*) 2 RC n 3 (D) RC n 6

Sol. In process of charging v = v0 (1 � e�t/) In process of discharging

vkos'ku ds nkSjku v = v0 (1 � e�t/) fujkos'ku ds nkSjku

t = n 0

0

V

V V

V = V0e

�t/

t = n 0V

V

For the considered situation

nh xbZ fLFkfr ds fy,



PHYSICS

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P1-15

Time taken during charging vkos'ku esa fy;k x;k le; t1 = 0 0

0 00 0

V Vn n

3V VV V

4 4

(from 0V

4 to 03V

4) =

4n 4 n n 3

3

( 0V

4ls 03V

4 rd ) =

4n 4 n n 3

3

Time taken during discharging fujkos'ku esa fy;k x;k le; t2 =

0

0

3V4n n 3V4

(from 03V

4 to 0V

4)

( 03V

4ls 0V

4 rd )

Total time period dqy vkorZdky = 2 n 3 = 2RC n 3

41. The graph of the current in the resistance R v/s charge of on the capacitor in one complete cycle is (Assume the voltmeter to be ideal.) [CP-EQ] (105)

,d pDdj ds nkSjku izfrjks/k R esa izokfgr /kkjk dk la/kkfj=k ij fLFkr vkos'k ds lkFk xzkQ crkb;sA

(ekuk oksYVehVj vkn'kZ gS)

VR

K2

K1

V0

+ �

(A*)

(B)



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P1-16

(C)

(D)

Sol. While charging vkos'ku d s nkSjku

+ �

+q

V0

R

i

�q

0q

V � iR 0C

0Vqi

RC R

When 0 0CV 3Vq i

4 4R

0 03CV Vq i

4 4R

While discharging fujkos'ku ds nkSjku

+ �

+q R

i

�q

q q

� � iR 0 i �C RC

When

0 0

0 0

3CV 3Vq i �

4 4RCV V

q i �4 4R



PHYSICS

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P1-17

Paragraph for Questions 42 and 43

iz'u 42 vkSj 43 d s fy, vuqPNsn

In a homogenous infinite solid cylindrical wire of radius R, there is a circular cylindrical and smooth cavity of radius R/2 whose axis is parallel to the conducting wire axis and displaced relative to it by

a distance R/2. A direct current of density J �k flows along the wire whose top view and side view

are shown in the figure. A point positive charge 'q' of mass m are given velocities v0�i , from point

C. Given that the value 2

0

8mvR

qJ

and all type of collision (if any) are assume perfectly elastic

collision.

R f=kT;k ds ,d leku vuUr Bksl csyukdkj rkj esa ;gk¡ ,d R/2 f=kT;k dh oÙ̀kkdkj csyukdkj fpduh xqgk

gSA ftldh v{k pkyd rkj dh v{k ds lekUrj gS rFkk ;g R/2 lkis{k nwjh ij gSA J �k ?kuRo dh fn"V /kkjk

rkj ds vuqfn'k izokfgr gS ft ldk 'kh"kZ n'̀; (top view) rFkk ik'oZ n'̀; (side view) fp=k esa n'kkZ;suqlkj gSA

m nzO;eku ds ,d /kukRed fcUnqor~ vkos'k 'q' dks fcUnq C ls osx v0�i fn;k tkrk gSA fn;k gqvk gS fd

2

0

8mvR

qJ

gS rFkk lHkh izdkj dh VDdj (;fn gS) dks iw.kZr% izR;kLFk VDdj ekusaA

[EM-FQ](104)

y

Axis of wire rkj dh v{k

Axis of cavity xqgk dh v{k

x

R/2

C

z

J O O �

C x

Read the above passage carefully and answer the following question.



PHYSICS

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P1-18

mijksDr vuqPNsn dks /;ku ls i<+s ,oa fuEu iz'uksa ds mÙkj nhft,A

42. The trajectory of motion of positive particle lies in plane : [EM-FQ](104)

/kukRed vkosf'kr d.k dh xfr ds iFk dk ry fdlesa fLFkr gS

(A) x-y plane ry esa (B) y-z plane ry esa (C*) x-z plane ry esa (D) can't say dg ugha ldrs

43. The radius of curvature of motion of positive charged particle is given by : [EM-FQ](104)

/kukRed vkosf'kr d.k dh xfr dh oØrk f=kT;k gS

(A) 0

0

2mv

q JR (B*) 0

0

4mv

q JR (C) 0

0

8mv

q JR (D) 0

0

mv

q JR

Sol. B = 2

)PCJ(

2

)POJ( 00

= 2

0 )]PCPO(J[

= 2

)COJ(0

=

4

JR0

O P C

00

JRB

4

(Uniform)

J

For positive charge r = 0

0

qB

mV =

0

4 mq RJ

/kukRed vkos'k ds fy, r = 0

0

qB

mV =

0

4 mq RJ

Negative charged particle moves along straight line _ .kkRed vkosf'kr d.k ljy js[kk ds vuqfn'k xfr'khy gSA

Paragraph for Questions 44 and 45

iz'u 44 vkSj 45 d s fy, vuqPNsn

Figure shows a Young's double slit experiment set-up. The source S of wavelength 4000 Å

oscillates along y-axis according to the equation y = sin t, where y is in millimeters and t is in seconds. The distance between two slits S1 and S2 is 0.5 mm. [YE-YA](104)



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P1-19

fp=k esa ,d ;ax f}fNnz iz;ksx O;oLFkk n'kkZ;h xbZ gSA 4000 Å rajxnS/;Z dk S L=kksr y-v{k ds vuqfn'k lehdj.k

y = sin t ds vuqlkj nksyu xfr djrk gS, tgk¡ y feyhehVj esa rFkk t lsd.M+ esa gSA nks fNnz S1 o S2 ds e/;

nwjh 0.5 mm gSA

y

S (0, 0) 0.5 mm

S2

S1

4m1m

x

P

44. The instant at which maximum intensity occurs at P for first time : [YE-YA](104)

P ij fdl {k.k ij vf/kdre rhozrk izFke ckj izsf{kr gksxh

(A) �11 59sin

160

(B) �11 27

sin80

(C*) �11 59

sin80

(D) �11 27

sin160

45. The instant at which minimum intensity occurs at P for the first time : [YE-YA](104)

P ij fdl {k.k ij U;wure rhozrk izFke ckj izsf{kr gksxh

(A) �11 59sin

160

(B*) �11 27

sin80

(C) �11 59

sin80

(D) �11 27

sin160

Sol. Path difference iFkkUrj

y

mean

S2

S1

0.25 mm

4m

x1 = d sin = 0.5 10�3 × 4

1025.0 3

= 32

10 6

metre

x2 = d sin = 0.5 × 10�3 ×

110y 3

= 210y 6

metre

For maximum intensity vf/kdre rhozrk ds fy, x1 + x2 = n

210y

3210 66

= 4000 × 10

�10

y = 8059



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P1-20

sin t = 8059

t =

8059

sin1 1

For minimum intensity U;wure rhozrk ds fy,

210y

3210 66

= 2000 × 10�10

y = 8027

t =

8027

sin1 1

For central maxima path difference is zero dsfUnz; mfPp"B ds fy, iFkkUrj 'kwU; gS d sin = � d sin

1y

4

y0

Position of central maximum is y0 = �4 sin t dsfUnz; mfPp"B dh fLFkfr y0 = �4 sin t gS

SECTION-3 : (Integer value correct Type) [k.M � 3 : (iw.kk±d eku lgh çdkj)

This section contains 5 questions. The answer to each question is a Two digit integer, ranging from 00 to 99 (both inclusive).

bl [k.M esa 5 ç'u gSaA çR;sd ç'u dk mÙkj 00 ls 99 rd ¼nksuksa 'kkfey½ ds chp dk nks vad ksa okyk iw.kk±d

gSA

Integer_(5)_(Double Digit)

46. A rubber band of natural length 1m (stiffness k = 22 Nm�1) is attached to a fixed point �A�, another

end is attached to a particle of mass 0.5 kg. The maximum downward velocity that can be given to the particle, so that it does not hit the floor is v m/s. Find v. (Assume Hooks law is valid even for longer deformation) [WE-UC](104)

1m izkd f̀rd yEckbZ dk ,d jcj csUM (cy fu;rkad k = 22 Nm�1) ,d fLFkj fcUnq �A� ls tqM+k gqvk gSA nwljk

fljk 0.5 kg nzO;eku ds ,d d.k ls tqM+k gqvk gSA d.k dks uhps dh vksj vf/kdre osx fn;k tk ldrk gS] og

v m/s gS rkfd ;g Q'kZ ls ugha VDdjk;sA v Kkr dhft,A (;g ekfu;s dh gqd dk fu;e vf/kd fod f̀r ds fy,

Hkh ykxw gksrk gSA)

A

2m

Floor

Ans. 02



PHYSICS

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P1-21

Sol. Using conservation of energy Å tkZ laj{k.k ls : 2max

1mv

2+ mg (2) = 21

k(1)2

on solving gy djus ij: vmax = 2 ms�1

47. Two block each of mass m are connected as shown in figure. Pulley are smooth and light. If the coefficient of static friction at all surfaces is s. If inclination is increased gradually then '' at

which the blocks begins to slide is given by 1 s100tan

p

. Find value of p. [FR-KF](104)

m nzO;eku ds nks CykWd fp=kkuqlkj tqM+s gq, gS f?kjuh ?k"kZ.kjfgr rFkk gYdh gSA ;fn lHkh lrgksa ij LFkSfrd

?k"kZ.k xq.kkad s gSA ;fn dks.k /khjs&/khjs c<+k;k tk;s rks dks.k ftl ij CykWd fQlyuk 'kq: dj nsxk]

1 s100tan

p

}kjk fn;k tkrk gSA p dk eku Kkr djksA

m m A

B

Fixed

Ans.(p = 20)

T

2T

mg sin � T � s mg cos = 0

T = mg sin � s mg cos

2T = mg sin + 3s mg cos



PHYSICS

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P1-22

2(mg sin � s mg cos ) = mg sin + 3 s mg cos

mg sin = 5s mg cos

tan = 5 s

= tan�1 (5s)



PHYSICS

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P1-23

48. The figure shows an RC circuit with a parallel plate capacitor. Before switching on the circuit, plate A of the capacitor has a charge �Q0 while plate B has no net charge. Now, at t = 0, the circuit is switched on. How much time (in second) will elapse before the net charge on plate A becomes

zero. (Given C = 1F, Q0 = 1mC, = 1000 V and 62 10

Rn3

)

[Modified by RGA Sir 2013-2014]

[CP-CK](104)

fp=k esa çnf'kZr RC ifjiFk esa lekUrj iê la/kkfj=k tqM+k gSA ifjiFk dks pkyw djus ds igys la/kkfj=k dh IysV A

ij �Q0 vkos'k gSA tcfd IysV B vukosf'kr gSA vc t = 0 ij ifjiFk pkyw fd;k tkrk gSA IysV A ij dqy

vkos'k 'kwU; gksus es fdruk le; (lSd.M esa) yxsxkA (fn;k gS% C = 1F, Q0 = 1mC, = 1000 V

rFkk62 10

Rn3

)

R

A B

�Q0

S

Ans. 02

Sol. Let at any time t charge flown through the plate B to plate A is q and instantaneous current is I.

ekuk fdlh Hkh le; t ij IysV B ls IysV A dh rjQ izokfgr vkos'k q gS rFkk rkR{k.khd /kkjk I gSA

I A B

�Q0+q

� q

R

I A B

�Q0+q

� q

R

From loop theorem ywi fl)kUr ls 02q Q

2C

+ R � = 0

Rdqdt

= 02q 2 C Q

2C

0

dq dt2 C Q 2q 2RC

Now for charge on plate A to be zero q = Q0.

vc IysV A ij vkos'k 'kwU; gS q = Q0.



PHYSICS

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P1-24

Integrating lekdyu djus ij 0Q t

00 0

dq dt2 C Q 2q 2RC

= t = RC n 0

0

2 C Q

2 C Q

Putting the value of C, Q0, and R We get t = 2 seconds.

C, Q0, rFkk R dk eku j[kus ij t = 2 lsd.M izkIr gksrk gSA

49. In the circuit shown in figure, the key k1, was closed for long time. At t = 0, key k1 is opened and k2

is closed. If the charge on capacitor C1 is N 3 C, at the instant, the energy stored in it is three

times of energy stored in inductor. Then N is : [CP-CK](104) [Made HRS sir 2012-13]

fp=k esa dqat h k1 yEcs le; ds fy;s cUn (on) j[kh xbZ gSA t = 0 ij dqath k1 dks [kqyk (off) rFkk k2 dks cUn

(on) dj fn;k x;k gSA ;fn la/kkfj=k C1 ij vkos'k N 3 C gSA rc ml {k.k ij blesa laxzfgr Å tkZ izsj.k

dq.Myh esa laxzfgr Å tkZ ls rhu xquh gSA rc N dk eku gksxkA

Ans. 10

Sol. Charge on Capacitor at t = 0 just after k2 closed Q0 = 20 C

t = 0 ij dqath k2 can djus ij la/kkfj=k ij vkos'k Q0 = 20 C

UC = 3UL and rFkk UC + UL = 20

1

Q2 C

2

0

1 1

QQ43 2C 2 C

0

3Q Q 10 3 c

2 Ans. (A)



PHYSICS

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P1-25

50. An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transitions. An electron is moving in an orbit of another hydrogen atom from which there can be a

maximum of three transitions. If ratio of the speed of the electron in these two orbits is ab

(in lowest

form), then find the value of a + b. Here a and b are integer :[Made AMG Sir_2013-14]

[MP-TR] (104)

,d bySDVªkWu gkbMªkstu ijek.kq dh ,d d{kk esa xfr'khy gS] tgkW ls og vf/kdre~ N% lØe.k dj ldrk gSA

,d bySDVªkWu vU; gkbMªkstu ijek.kq dh ,d d{kk eas xfr'khy gS] tgkW ls og vf/kdre~ rhu laØe.k dj ldrk

gSA bu nks d{kkvksa esa bySDVªkWu dh pky dk vuqikr ab

(U;wure~ :i esa) gSA rc, a + b dk eku Kkr djksA ;gk¡

a o b iw.kkZad gS %

Ans. 7

Sol. a = 4

b = 3 a + b = 7



Page # 1

Course JPT-1 (JEE ADVANCE) Test Date : 10-05-2015

Test Type : JEE ADVANCED (ELPD)

SYLLABUS :

FULL SYLLABUS

Test Pattern :


1 to 12 SCQ 12 3 �1 36



26 to 37 SCQ 12 3 0 36



51 to 62 SCQ 12 3 0 36



75 225


1 to 15 MCQ 15 4 0 60

16 to 19 Match matrix listing 4 3 �1 12

20 to 34 MCQ 15 4 0 60


39 to 53 MCQ 15 4 0 60


57 216

Paper-1

Paper-2

Total Total

TotalTotal

Maths

Physics

Chemistry

Physical Inorganic Chemistry Paper-1 Organic Chemistry Paper-1 SCQ (8) SCQ (4) Comp.(4 x 2Q.) (2) Comp.(4 x 2Q.) (2) Integer (Double digit) (3) Integer (Double digit) (2) Physical Inorganic Chemistry Paper-2 Organic Chemistry Paper-2 MCQ (10) MCQ (5) Match Listing type (2) Match Listing type (2)

Page # 2

JEE (ADVANCED) CHEMISTRY PAPER SKELETON

Faculty Name : Test Name : JR (JPT-1) Faculty preparing the TEST PAPER should fill it according to paper pattern and submit it with finalisation of paper at SMD.

PAPER-1

S. No. TYPE (P) (I) (O) TOPIC(S) SUBTOPIC(S)

DIFFICULTY LEVEL : Easy (E), Moderate (M), Tough (T)

Question Nature

Is the solution there (1/0)

51 SCQ (I) COR COR-ACFT M 206 1

52 SCQ (I) DBC DBC-LA E 203 1

53 SCQ (P) CHK CHK-PRFO T 206 1

54 SCQ (P) ECH ECH-ANE M 205 1

55 SCQ (P) SST SST-BSS E 201 1

56 SCQ (P) GST GST-DLPP M 207 1

57 SCQ (P) CBO CBO-HYB M 206 1

58 SCQ (P) TDS TDS-HIE T 206 1

59 SCQ (O) CAD CAD-AEA M 203 1

60 SCQ (O) SIM SIM-GI M 203 1

61 SCQ (O) AK AK-NAR M 204 1

62 SCQ (O) HYC HYC-EARO M 203 1

63 Comp. 1 (Q.1) (P) CHK CHK-ZFO M 206 1

64 Comp. 1 (Q.2) (P) CHK CHK-ZFO M 206 1

65 Comp. 2 (Q.1) (I) QUA QUA-DG M 206 1

66 Comp. 2 (Q.2) (I) QUA QUA-DG M 206 1

67 Comp. 3 (Q.1) (O) SIP SIP-TFG M 204 1

68 Comp. 3 (Q.2) (O) SIP SIP-TFG M 204 1

69 Comp. 4 (Q.1) (O) AK AK-NAR M 204 0

Page # 3

70 Comp. 4 (Q.2) (O) AK AK-NAR M 204 1

71 Double Integer

type (I) QUA QUA-VNVIG M 203 1

72 Double Integer

type (I) PBC PBC-XV M 203 1

73 Double Integer

type (I) QUA QUA-CG T 206 1

74 Double Integer

type (O) AK AK-GR M 204 1

75 Double Integer

type (O) AC AC-PH M 204 1

Page # 4

Physical Inorganic Chemistry Paper-1 SCQ (8) 51. Which of the following is suitable match : (COR-ACFT_206(I)) Complex Colour (1) M (L1)6 (p) Red (2) M (L2)6 (q) Yellow (3) M (L3)6 (r) Green (4) M (L4)6 (s) Violet Strength of ligand L2 > L3 > L4 > L1 (use concept of complimentary colour)

fuEu esa ls dkSu lgh lqesfyr gS & ladqy jax

(1) M (L1)6 (p) yky

(2) M (L2)6 (q) ihyk (3) M (L3)6 (r) gjk (4) M (L4)6 (s) cSaxuh fyxs.M dh lkeF;Z L2 > L3 > L4 > L1 (lEiwjd jax ds fl)kUr dk iz;ksx djsa)

(1) (2) (3) (4) (1) (2) (3) (4)

(A) (r) (q) (s) (p) (B*) (r) (q) (p) (s)

(C) (p) (s) (q) (r) (D) (s) (r) (p) (q)

Ans. (A-r) ; (B-q) ; (C-p) ; (D-s)

Sol.

B

OY

G

R

V

absorbed � L2 < L3 < L4 < L1

M(L2)6 Yellow

M(L3)6 Red

M(L4)6 Violet

M(L1)6 Green

gy%

B

OY

G

R

V

vo'kksf"kr � L2 < L3 < L4 < L1

M(L2)6 ihyk

M(L3)6 yky

M(L4)6 cSaxuh

M(L1)6 gjk 52. True & False : (DBC-LA_203(I)) (1) Ce+4 is an oxidizing agent & colourless (2) Lu3+ is colourless (3) Actinoids exhibit higher number of oxidation states than lanthanoid. (4) In photography, hypo is complexing agent as well as reducing agent. (5) All 3d elements give H2 with 1 M HCl

Page # 5

lR; rFkk vlR; & (1) Ce+4 vkWDlhdkjd rFkk jaxghu gksrk gSA (2) Lu3+ jaxghu gksrk gSA (3) ,fDVukWbM] ysUFksukWbMks dh vis{kk vkWDlhdj.k voLFkk dh mPp la[;k n'kkZrs gSA (4) QksVksxzkQh esa gkbiks ,d ladqy vfHkdeZd (complexing agent) rFkk vipk;d gksrk gSA (5) leLr 3d rRo 1M HCl ds lkFk H2 nsrs gSaA (A) F T T F T (B) T F T F T (C*) T T T F F (D) F F T F F Sol. For (1), (2), (3) refer class notes (4) Hypo is complexing agent only in photography (5) Cu does not give H2 with 1 M HCl. (1), (2), (3) ds fy, uksV~l ns[ksaA (4) gkbiksdsoy QksVksxzkQh esa ladqy vfHkdeZd gksrk gSA (5) Cu 1 M HCl ds lkFk H2 ugha nsrk gSA

53. A �K 2sec

2B (CHK-PRFO_206(P))

A �K 4sec

3C A is undergoing two reactions as given above. Final mol % of B in mixture when 40% of A is consumed : A mijksDr n'kkZ;h x;h vfHkfØ ;k;sa nsrk gSA feJ.k esa B dk vfUre eksy % crkb,] tc 40% A dk mi;ksx gks pqdk

gS& (A) 9 % (B) 29% (C) 10% (D*) 16%

Sol.

A 2B 2x

3C 3y

100�x�y

[B][C]

= 1

2

2K

3K =

13

nA = 60 nB + nC = 2x + 3y x + y = 40

2x3y

= 13

2x = y

x = 403

y = 803

Mole % of B B dk eksy % = 2x

2x 3y 60 × 100 =

803

8080 60

3

× 100 = 16%

54. An electrode of a cell is made by dipping Ag rod in 0.1 M, 1L KCl solution after some time EAg

+ / Ag = 0.32 V

if o

Ag /AgE

= 0.8 V then milimoles of AgCl formed is : (Ksp of AgCl = 10�10) (Take 2.303RT

F = 0.06)

(ECH-ANE_205(P)) ,d lSy ds bysDVªkWM dks 0.1 M, 1L KCl foy;u esa Ag NM+ dks Mwcksdj cuk;k tkrk gS dqN le; ds i'pkr~ EAg

+ / Ag =

0.32 V ;fn o

Ag /AgE

= 0.8 V gS] rc fufeZr AgCl ds feyheksy gS & (AgCl dk Ksp = 10�10) (2.303RT

F = 0.06

yhft;s) (A) 70 (B) 80 (C*) 90 (D) 100 Sol. Ag+ + e� Ag

Page # 6

0.32 = 0.8 � 0.06

1log

1

[Ag ]

[Ag+] = 10�8 M [Cl�] = 10�2 M �Cl

n = 0.01

Consumed mole of Cl� = 0.1 � 0.01 = 0.09 Formed milimoles of AgCl = 90 Sol. Ag+ + e� Ag

0.32 = 0.8 � 0.06

1log

1

[Ag ]

[Ag+] = 10�8 M [Cl�] = 10�2 M �Cl

n = 0.01

Cl� ds dke esa fy, x;s eksy = 0.1 � 0.01 = 0.09 AgCl ds fufeZr feyh eksy = 90 55. In which of following at least one of angle is 90 and at least two axial lengths are same ? (SST-BSS_201(P)) (A) Monoclinic (B) Rhombohedral (C) Orthorhombic (D*) Tetragonal fuEu esa ls fdlesa de&ls&de ,d dks.k 90 gS rFkk de&ls&de nks v{kh; yEckbZ;k¡ leku gS\ (A) ,durk{k (B) f=kleurk{k (C) fo"keyEck{k (D*) f}leyEck{k Sol. Factual rF;kRed 56. A 20 litre rigid cylinder containing N2 at 20 atm and 300 K is placed in big jar containing air at 1 atm and

300 K and finally sealed as given below. Now N2(g) was allowed to leak into to jar at constant temperature. Final total pressure in jar is 11 atm. Final volume of jar in litre.

(GST-DLPP_207(P)) 20 atm rFkk 300 K ij N2 ;qDr ,d 20 yhVj n<̀+ flys.Mj 1 atm rFkk 300 K ij ok;q ;qDr ,d cM+s tkj esa j[kk gqvk

gS rFkk vUr esa bls uhps n'kkZ;s vuqlkj cUn dj nsrs gSA vc N2(g) dk fu;r rki ij tkj esa ls fu%ljhr gksus fn;k tkrk gSA tkj esa vfUre dqy nkc 11 atm gSA tkj d k vfUre vk;ru yhVj esa crkb,A

Cylinder

Jar

(A) 18 litre (B) 40 litre (C) 58 litre (D*) 38 litre Sol. nT =

2Nn in cylinder + airn in Jar

11(V + 20) = (20 × 20) + 1 × V V = 18 litre Volume of Jar = V + 20 = 38 litre gy% nT =

2Nn flys.Mj esa + nok;q tkj esa

11(V + 20) = (20 × 20) + 1 × V V = 18 yhVj tkj dk vk;ru = V + 20 = 38 yhVj 57. In how many of following molecules, all atoms are in same plane? (CBO-HYB_206(I)) fuEu esa ls fdrus v.kqvksa esa lHkh ijek.kq leku ry esa mifLFkr gS\ ClF3 H2O PCl3 BF3

SF4 H2S OCl2 SO3

XeF6 NH3 C6H6 XeF2

XeF4 PCl5 l2Cl6 PH3

Page # 7

(A) 12 (B) 0 (C*) 10 (D) 11 Sol. Planar molecules � ClF3, XeF4, H2O, H2S, OCl2, C6H6, I2Cl6, BF3, SO3, XeF2

leryh; v.kq & ClF3, XeF4, H2O, H2S, OCl2, C6H6, I2Cl6, BF3, SO3, XeF2 58. For polytropic process a diatomic gas has following curve of P & V.

Then molar heat capacity for this process will be � (TDS-HIE_206(P))

,d f}ijekf.kd xSl ds ikWfyVªksfid izØe ds fy, fuEu P o V oØ izkIr gksrk gS &

rc bl izØe ds fy, eksyj Å "ek /kkfjrk D;k gksxh \

(2, 4) P

45º

V

(A) 3.5 R (B*) 4.5 R (C) 2.5 R (D) 4.3 R

Sol. Since pwafd dvdp

= �x

vP

(for polytrophic process PVx = const) ¼ikWfyVªksfid izØe ds fy, PVx = fu;r½

�1 = �x ×

24

x = 12

CM = CV + x1

R

( CV = R25

)

CM = R25

+ R

11�

2

= 2.5 R + 2 R = 4.5 R

59. H3CCOC2H5 + CH3COOC2H5 2 5

3

C H ONa

H O (P) Major

(CAD-AEA(O)(M)_203) (A*) (P) Predominantly exists in enolic from. (B) (P) Predominantly exists in keto from. (C) (P) When reacted with C2H5OD + C2H5ONa, all the hydrogens will be replaced by deuterium. (D) (P) Will give coloured precipitate with neutral FeCl3 but will not give test with 2,4-DNP.

H3CCOC2H5 + CH3COOC2H5 2 5

3

C H ONa

H O (P) eq[;

(A*) (P) izeq[k :i ls bZukfyd :i esa jgrk gSA (B) (P) izeq[k :i ls fdVks :i esa jgrk gSA (C) ;kSfxd (P) tc C2H5OD + C2H5ONa ds lkFk fØ;k djrk gS rks lHkh gkbMªkstu ijek.kq M;wVsfj;e }kjk izfrLFkkfir

gksxsA (D) ;kSfxd (P) mnklhu FeCl3 ds lkFk fØ;k djds jaxhu vo{ksi nsxk ysfdu 2,4-DNP ds lkFk fØ;k ugha d jsxkA

Page # 8

Sol. H5C2COCH3 2 5C H ONa

A / B

CH3�C�OC2H5 || O

CH3�CH2�C�CH2�C�CH3 || O

|| O

H5C2COCH2Na + C2H5OH

60. Which of the following oximes is expected to exhibit optical isomerism and geometrical isomerism both. fuEu esa ls dkSulk vkWDlhe izdkf'kd leko;ork o T;kfefr; leko;ork nksuksa n'kkZrk gS %& ( (SIM-GI(O)(M)_203)

(A)

C=N�OH H5C6

H3C

(B)

C=N�OH

H5C6

(C)

C=N�OH

H5C6

(D*)

N�OH

CH3

Sol. A, B, C will show only G.I. However D also have one chiral carbon so will show G.I. & O. I. both. Sol. A, B, C dsoy T;kfefr; leko;ork n'kkZ;sxsA ;|fi D ,d fdjSy dkcZu j[krk gS blfy, ;g T;kfefr; o

izdkf'kd nksuksa leko;ork n'kkZ;sxkA

61.

OH

CH2�CH=O 2 4

NaCN

H SO(95%)

(P) Major eq[;

Major product (P) will be : eq[; mRikn (P) gksxk% (AK-NAR(O)(M)_204)

(A)

OH

CH2�CH�CN | OH

(B)

OH

CH2�CH�COOH| OH

(C*)

O O

(D)

OH

O O

Sol.

OH

CH2�CH=O 2 4

NaCN

H SO(95%)

OH

CH2�CH�COOH| OH

2�2H O

O O

62.

P

(1) Hg(OAc)2H2O

(2) NaBH4 (R)

(2) H2O2 + OH NaBH

(1) BH3 + THF (S)

(Q) H3O

Find major product Q, R & S respectively. (HYC-EARO(O)(M)_203)

(A*)

OH ,

OH and

OH

(B)

OH ,

OH and

OH

Page # 9

(C)

OH ,

OH

and

OH

(D)

OH

all of them

P

(1) Hg(OAc)2H2O

(2) NaBH4 (R)

(2) H2O2 + OH NaBH

(1) BH3 + THF (S)

(Q) H3O

eq[; mRikn Q, R o S Øe'k% gS %

(A*)

OH ,

OH o

OH

(B)

OH ,

OH o

OH

(C)

OH ,

OH

o

OH

(D)

OH

lHkh

Sol. Q is formed by addition of water by M.K. rule with rearrangement however R is formed by addition of H2O

without rearrangement. Product (S) is formed by addition of water by anti M.K. rule. Sol. ;kSfxd Q iquZfoU;kl gksus ij ekjdksuhdkWQ fu;e ds vk/kkj ij ty ds ;ksx }kjk curk gSA tcfd ;kSfxd R fcuk

iquZfoU;kl ds H2O ds ;ksx }kjk curk gSA mRikn (S) ,UVhekjdksuhdkWQ fu;e ls ty ds ;ksx }kjk fufeZr gksrk gSA Comp.(4 x 2Q.) (2)

Paragraph for Question Nos. 63 to 64 iz'u 63 ls 64 d s fy, vuqPNsn

A & B are volatile liquids with PA

0 = 340 mm and PB0 = 420 mm Hg. To a flask containing 8 moles of A, 5

moles of B were added. As soon as B is added, A starts polymerizing forming Ax (solid) which is soluble in both liquids A and B.

Polymerisation of A follows 1st order kinetics. Vapour pressure of solution at time is 300 mm Hg. Partial vapour pressure of B is 210 mm after 1 hr. (CHK-ZFO_206(P)) A rFkk B, PA

0 = 340 mm rFkk PB0 = 420 mm Hg ;qDr ok"i'khy nzo gSA 8 eksy A okys ¶ykLd esa 5 eksy B feyk;k

x;kA tSls gh B feykrs gS] A cgqydhd r̀ gksdj Ax(Bksl ) cukuk izkjEHk djrk gSA tks nksuksa nzoksa A rFkk B esa ?kqyu'khy gSA A dk

cgqydhdj.k izFke dksfV cy xfr ds vuqlkj gksrk gSA le; ij foy;u dk ok"i nkc 300 mm Hg gSA 1 ?kUVs ds i'pkr~ B dk vkaf'kd ok"inkc 210 mm Hg gSA

63. What is value of x in Ax (Solid)? (CHK-ZFO_206(P)) Ax(Bksl) esa x dk eku D;k gS\ (A) 1 (B) 2 (C) 3 (D*) 4 64. Rate constant in per hour for association of A is : (CHK-ZFO_206(P)) A ds laxq.ku ds fy, izfr ?kUVs nj fu;rkad gSa & (A) 2 ln 2 (B*) ln 2 (C) ln 2/4 (D) ln2/4 Sol. (63) & (64) xA Ax t = 0 8 mol 0

1 hr 8 � a ax

8x

Page # 10

420 � 300

300 =

8 / x5

x = 4 210 = 420 XB

XB = 12

= 5

a5 8 � a

4

13 � 3a4

= 10

3a4

= 3

a = 4

k = 11n

84

= n2


iz'u 65 ls 66 d s fy, vuqPNsn

A solid (A) is water insoluble, while a red solid (B) is water soluble. Heating (A) and (B) produces a colorless gas (C) and a reddish brown gas (D) respectively, leaving residues (E, yellow) and (F, black). (A) and (B) when heated together produce a green colored solid (G). (G) is also obtained on heating (E) and (F) together. (E) becomes white on adding water, but it fails to dissolve in water. Then

(QUA-DG_206(I)) ,d Bksl (A) ty esa vfoys; gS tcfd ,d yky Bksl (B) ty esa foys; gSA (A) rFkk (B) dks xeZ djus ij] Øe'k% jaxghu

xSl (C) rFkk yky Hkwjh xSl (D) nsrs gS rFkk lkFk gh Øe'k% vof'k"V (E, ihyk) rFkk (F, dkyk) nsrs gSA tc (A) rFkk (B) nksuksa dks ,d lkFk xeZ fd;k tkrk gS rc gjs jax dk Bksl (G) curk gSA tc (E) rFkk (F) dks ,d lkFk xeZ djrs gS rc Hkh (G) izkIr fd;k tk ldrk gSA (E) esa ty feykus ij ;g lQsn gks tkrk gS ijUrq ;g ty esa foys; ugha gksrk gSA rc

65. (A) and (B) are respectively (QUA-DG_206(I)) (A) ZnCO3 and Pb3O4 (B) CrO3 and ZnSO4 (C) CoCO3 and ZnSO4 (D*) ZnCO3 and Co(NO3)2

(A) rFkk (B) Øe'k% gSa % (A) ZnCO3 rFkk Pb3O4 (B) CrO3 rFkk ZnSO4

(C) CoCO3 rFkk ZnSO4 (D*) ZnCO3 rFkk Co(NO3)2 66. When (E) and (F) react to give (G) (QUA-DG_206(I)) (A*) (E) functions as an acid (B) (E) functions as an base (C) (E) functions as an oxidant (D) (E) functions as a reductant tc (E) rFkk (F) fØ;k djds (G) nsrs gS rc &

(A*) (E) ,d vEy ds leku dk;Z djrk gSSA (B) (E) ,d {kkj ds leku dk;Z djrk gSSA (C) (E) ,d vkWDlhdkjh ds leku dk;Z djrk gSSA (D) (E) ,d vipk;ddkjh ds leku dk;Z djrk gSS Sol. (65) & (66)

3 2ZnCO ZnO CO

(A) (E) (C)

;

1Co(NO ) 2NOCoO O3 2 2 22(F) (D)(B), red

Note : ZnO is yellow when hot, white when cold. fVIi.kh : ZnO xeZ djus ij ihyk gks tkrk gS tcfd B.Mk djus ij lQsn gksrk gSA

CoZnOCoO ZnO (green)2

(F) (E) (G)

¼gjk½


iz'u 67 ls 68 d s fy, vuqPNsn On the bases of following information answer the given two questions.

Page # 11

C5H10O

(P)

dil.H2SO4 Q + R

(1) I2+NaOH (2) H3O

CHI3 + S

R reduces Tollens reagent

(1) I2+NaOH (2) H3O

CHI3 + T fuEu lwpukvksa ds vk/kkj ij fn;s x;s nks iz'uks ds mÙkj fnft,

C5H10O

(P)

ruq.H2SO4 Q + R

(1) I2+NaOH (2) H3O

CHI3 + S

R vipf;r VkWysu vfHkdeZd

(1) I2+NaOH (2) H3O

CHI3 + T (SIP-TFG(O)(M)_204) 67. P will be � P gksxk & (SIP-TFG(O)(M)_204) (A) CH2=C�O�CH2�CH3

CH3

(B) CH2=CH�O�CH2�CH2�CH3

(C*) CH2=CH�O�CH�CH3

CH3

(D) CH2=CH�CH2�O�CH2�CH3

68. Which amongs the following is incorrect statement � (SIP-TFG(O)(M)_204) (A) [Q] is also formed when propene is reacted with dil.H2SO4. (B*) [Q] will also give coloured complex with neutral FeCl3. (C) The product formed by dry distillation of calcium salt of [S] will give acetone. (D) [T] will also give silver mirror with ammonical solution of AgNO3. fuEu esa ls dkSulk dFku xyr gS % (A) ;kSfxd [Q] ruq H2SO4 ds lkFk izksihu dh fØ;k }kjk Hkh fufeZr gksrk gSA (B*) ;kSfxd [Q] mnklhu FeCl3 ds lkFk jaxhu foy;u cukrk gSA (C) ;kSfxd [S] ds dSfYl;e yo.k ds 'kq"d vklou }kjk fufeZr mRikn ,lhVksu nsxkA (D) ;kSfxd [T], AgNO3 ds veksuhÑr foy;u ds lkFk jtr niZ.k Hkh nsxkA Sol. (67) & (68)

CH2=CH�O�CH�CH3

(P)

dil.H2SO4

(Q) (R)

I2+NaOH

CHI3 + HCOOH (T)

CH3

CH3�CH�OH

CH3

+ CH2=CH�OH CH3�CH=O

CH2�CH�OH

(Q)

I2 + NaOH

(S) CH3

CHI3 + CH3COOH Ca(OH)2 CH3�CO�CH3


P 3

2

O

Zn H O Q + R

Q conc. NaOH

Benzylalcohol + Salt of Benzoic acid

R dil.NaOH

Ph�C�CH2�C�Ph

CH3

OH O

NaCN

HCl 3H O

S

Page # 12

Based on above reaction sequence, answer the following questions.

iz'u 69 ls 70 d s fy, vuqPNsn

P 3

2

O

Zn H O Q + R

Q NaOH

lkUnz csfUty ,Ydksgy + csUtksbd vEy dk yo.k

R NaOH

ruq

Ph�C�CH2�C�Ph

CH3

OH O

NaCN

HCl 3H O

S

mijksDr vfHkfØ;k vuqØe ds vk/kkj ij fuEu iz'uksa ds mÙkj fnft, %!" (AK-NAR(O)(M)_204) 69. Structure of S is � ;kSfxd S dh lajpuk gS % (AK-NAR(O)(M)_204)

(A)

O

Ph OH

Ph H3C O

(B*)

O

HO

H3C

Ph Ph

O (C)

O

C�OH H3C

Ph Ph

O (D)

O

HO H3C Ph Ph

O

70. Which is not correct for P ? (AK-NAR(O)(M)_204) (A) P will have two diastereomers. (B) P is also prepared by dehydration of 1,2-Diphenyl propane-2-ol. (C) Index of hydrogen deficiency of P will be 9. (D*) One of the ozonolysis product of P will not give silver mirror with ammonical AgNO3 (Tollen's reagent) P ds fy, dkSulk xyr gS \ (A) P ds nks foofje leko;oh gksxsaA (B) P dks 1,2-MkbZQsfuy izksisu-2-vkWy ds futZyhdj.k }kjk Hkh cuk;k tk ldrk gSA (C) ;kSfxd P dk gkbMªkstu U;wurk funsZ'kkad 9 gksxkA (D*) ;kSfxd P dk dksbZ ,d vkstksuhvi?kVu mRikn veksfuÑr AgNO3 (VkWysu vfHkdeZd ) ds lkFk jtr niZ.k ugha nsxkA Sol. P must be Ph�C=CH�Ph

CH3 Sol. P Ph�C=CH�Ph

CH3

gksxkA

Integer (Double digit) (3)

71. How many of following give pink precipitate with NaOH (precipitate may or may not dissolve with excess of NaOH)? (QUA-VNVIG_203(I))

fuEu esa ls fdrus vk;u NaOH ds lkFk xqykch vo{ksi nsrs gS ¼vo{ksi.k NaOH ds vkf/kD; ds lkFk fo;ksftr gks Hkh ldrk gS ;k ugh Hkh gks ldrk gS½\

Ag+, Pb2+, 22Hg , Bi3+, Cu2+, Cd2+, Cr3+, Mn2+, Co2+, Ni2+, Ba2+, Ca2+, Mg2+

Ans. 01 Sol. Co2+, Co(OH)2 is pink precipitate. Co2+, Co(OH)2 xqykch vo{ksi gSA 72. How many of following will produce atleast one coloured gas on heating? (PBC-XV_203(I)) fuEu esa ls fdrus ;kSfxd xeZ djus ij de&ls&de ,d jaxhu xSl nsrs gSa\ FeSO4, LiNO3, KNO3, Co(NO3)2, H3BO3, K2Cr2O7, KMnO4, NH4NO3, NH4NO2, (NH4)2CO3, AlCl3.6H2O,

NaN3, BI3 Ans. 03 Sol. Co(NO3)2, LiNO3, BI3

Page # 13

73. H3PO4 + HNO3 + (NH4)2MoO4 Oxidation number of Mo in product is x.

Cr2O72� + H+ + NaCl Oxidation state of Cr in product is y

SCN� + MnO2 Number of bonds in sulphur containing product is z. Find x + y + z/ (QUA-CG_206(I))

H3PO4 + HNO3 + (NH4)2MoO4 mRikn esa Mo dh vkWDlhdj.k la[;k x gSA

Cr2O72� + H+ + NaCl mRikn esa Cr dh vkWDlhdj.k la[;k y gSA

SCN� + MnO2 mRikn ;qDr lYQj esa ca/kksa dh la[;k z gSA x + y + z Kkr dhft,A (QUA-CG_206(I)) Ans. 16 x = +6 (NH4)3PMo12O40 y = +6 CrO2Cl2

z = 4 Z NC�S�S�CN

74.

CH=O

3

3

(1) CH MgBr ROR

(2) H O

[P] 2 4conc.H SO

[R] 2H Ni

[S]

Calculated total number of monochloro products when [S] is heated with Cl2.

CH=O

3

3

(1) CH MgBr ROR

(2) H O

[P] 2 4H SO

lkUnz [R] 2H Ni [S]

;kSfxd [S] dks tc Cl2 ds lkFk xeZ fd;k t krk gS rks cuus okys dqy ,dy Dyksjks mRikn crkb;sA (Made by IKS Sir On April 2015) (AK-GR(O)(M)_204) Ans. 20

Sol.

CH=O

3CH MgBr

CH�CH3

OH

(P)

H

(R) 2H

(S)

(S) 2Cl

2

Cl

4 2

+

2

Cl

4 2

+

Cl

4 isomer

pseudochiral

+

2 Cl 8 3

Sol.

CH=O

3CH MgBr

CH�CH3

OH

(P)

H

(R) 2H

(S)

(S) 2Cl

2

Cl

4 2

+

2

Cl

4 2

+

Cl

4 leko;oh

Nnefdjsy

+

2 Cl 8 3

75. An organic compound (P) on treatment with CHCl3 & KOH gives (Q) & (R) both of which gives same

compound (S) when distilled with Zn. Oxidation of (S) gives aromatic compound (T) of formula C7H6O2. The sodium salt of (T) with sodalime gives (X) which can also be obtained by distilling (P) with Zn powder.

Find molecular weight of (P) ? ,d dkcZfud ;kSfxd (P) CHCl3 o KOH ds lkFk vfHkfØ;k ij (Q) o (R) nsrk gSA nksuks ;kSfxdks dks tc Zn ds lkFk

vklfor fd;k tkrk gS rks leku ;kSfxd (S) cukrs gSA ;kSfxd (S) vkWDlhdj.k ij v.kqlw=k C7H6O2 okyk ,sjkseSfVd ;kSfxd

(T) cukrk gSA ;kSfxd (T) dk lksfM;e yo.k lksMk ykbe ds lkFk fØ;k dj ;kSfxd (X) cukrk gS ftls Zn pw.kZ ds lkFk ;kSfxd (P) ds vklou }kjk Hkh izkIr fd;k tk ldrk gSA ;kSfxd (P) dk v.kqHkkj Kkr fdft, \

(AC-PH(O)(M)_204)

Page # 14

Ans. 94

Sol.

OH

OH CH=O

(Q)

+ para (R)

Zn

CH=O

(S)

(X)

CHCl3 + KOH

Zn +

(T)

NaOH CaO

COOH Oxidation (P)

JEE Advanced : MOCK TEST PAPER 2016

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Transcript of JEE Advanced : MOCK TEST PAPER 2016